Write an equation in the indicated form for each statement described below.
(A graph shows two lines intersecting at x=-2 and y= 1 in a tilted X shape)
A. Write an equation in slope-intercept form for the line parallel to the red line and going through the point (3,5).
B. Write an equation in slope-intercept form for the line perpendicular to the blue line and going through the point (-4, 0).
0.6 in indicated form
A. To find the equation of a line parallel to the red line and going through the point (3,5), we need to find the slope of the red line first.
Looking at the graph, we can see that the red line has a slope of -1. This means that any line parallel to the red line will also have a slope of -1.
Using the point-slope form of a linear equation, which is y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line, we can write the equation for the line parallel to the red line and passing through the point (3,5) as:
y - 5 = -1(x - 3)
Expanding this equation gives us the slope-intercept form:
y - 5 = -x + 3
Simplifying further, we get:
y = -x + 8
So, the equation in slope-intercept form for the line parallel to the red line and going through the point (3,5) is y = -x + 8.
B. To find the equation of a line perpendicular to the blue line and going through the point (-4,0), we need to find the negative reciprocal of the slope of the blue line.
Looking at the graph, we can see that the blue line has a slope of 1. This means that any line perpendicular to the blue line will have a slope of -1/1 or -1.
Using the point-slope form again, we can write the equation for the line perpendicular to the blue line and passing through the point (-4,0) as:
y - 0 = -1(x - (-4))
Simplifying this equation gives us:
y = -x - 4
So, the equation in slope-intercept form for the line perpendicular to the blue line and going through the point (-4,0) is y = -x - 4.
A. To find the equation of a line parallel to the red line and passing through the point (3,5), we need to find the slope of the red line first. Looking at the graph, we can see that the red line has a slope of 1 (since it goes up by 1 unit for every 1 unit it moves to the right). So, a line parallel to the red line will also have a slope of 1.
Using the point-slope form of a line, which is given by y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line, we can substitute the values into the equation:
y - 5 = 1(x - 3)
Simplifying the equation, we get:
y - 5 = x - 3
Now, let's isolate y:
y = x - 3 + 5
Finally, simplifying further, we get:
y = x + 2
Therefore, the equation of the line parallel to the red line and passing through the point (3,5) is y = x + 2.
B. To find the equation of a line perpendicular to the blue line and passing through the point (-4, 0), we need to find the negative reciprocal of the slope of the blue line. Looking at the graph, we can see that the blue line has a slope of -1/2 (since it goes down by 1 unit for every 2 units it moves to the right). So, the negative reciprocal of -1/2 is 2.
Using the point-slope form of a line, the equation becomes:
y - 0 = 2(x - (-4))
Simplifying the equation, we get:
y = 2(x + 4)
Expanding further, we get:
y = 2x + 8
Therefore, the equation of the line perpendicular to the blue line and passing through the point (-4, 0) is y = 2x + 8.
A: If the red line has slope m1, then your line is
y-5 = m1(x-3)
B: If the blue line has slope m2 then your line is
y = (-1/m2)(x+4)
So, use the needed slopes, and massage the equations into the form
y = mx+b